A string that is stretched between fixed supports separated by 75.0 cm has resonant frequencies of 420 and 315 Hz, with no intermediate resonant frequencies. What is the lowest resonant frequency?

Options

A. 250 Hz
B. 317 Hz
C. 180 Hz
D. 105 Hz

Answer

To solve the problem of finding the lowest resonant frequency of a string stretched between fixed supports, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resonant Frequencies**:
The resonant frequencies of a string fixed at both ends can be described by the formula:
\[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f_n \) is the resonant frequency,
- \( n \) is the harmonic number (1 for the fundamental frequency, 2 for the first overtone, etc.),
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

2. **Given Frequencies**:
We are given two resonant frequencies: \( f_1 = 420 \, \text{Hz} \) and \( f_2 = 315 \, \text{Hz} \). Since there are no intermediate resonant frequencies, these correspond to consecutive harmonics.

3. **Identifying Harmonic Numbers**:
Let's denote the harmonic number corresponding to \( f_1 \) as \( n \) and the harmonic number corresponding to \( f_2 \) as \( n-1 \). Thus:
\[ f_1 = f_n = 420 \, \text{Hz} \quad \text{and} \quad f_2 = f_{n-1} = 315 \, \text{Hz} \]

4. **Setting Up the Ratio**:
From the formula for resonant frequencies, we can set up the ratio:
\[ \frac{f_1}{f_2} = \frac{n}{n-1} \]
Plugging in the values:
\[ \frac{420}{315} = \frac{n}{n-1} \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 420(n-1) = 315n \]
Simplifying this:
\[ 420n - 420 = 315n \]
\[ 420n - 315n = 420 \]
\[ 105n = 420 \]
\[ n = \frac{420}{105} = 4 \]

6. **Finding the Lowest Frequency**:
Now that we have \( n = 4 \), we can find the lowest resonant frequency, which corresponds to \( n = 1 \):
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Using the relationship between the frequencies:
\[ f_n = \frac{n}{n-1} f_{n-1} \]
We can find the fundamental frequency \( f_0 \):
\[ f_4 = 420 \, \text{Hz} \quad \text{and} \quad f_3 = 315 \, \text{Hz} \]
The fundamental frequency can be calculated as:
\[ f_0 = \frac{f_4}{4} = \frac{420}{4} = 105 \, \text{Hz} \]

### Final Answer:
The lowest resonant frequency is:
\[ \boxed{105 \, \text{Hz}} \]